Question

A library has a copies of one book, b copies each of two books, c copies each of three books, an single copy of d books. The total number of ways in which these books can be arranged in a shelf is equal to

• A. $\frac{(a+2b+3c+d)!}{a!(b!{)}^2(c!{)}^3}$
• B. $\frac{(a+2b+3c+d)!}{a!(2b!)(c!{)}^3}$
• C. $\frac{(a+b+3c+d)!}{(c!{)}^3}$
• D. $\frac{(a+2b+3c+d)!}{a!(2b!)(3c!)}$

Answer 1

The correct option is A $\frac{(a+2b+3c+d)!}{a!(b!{)}^2(c!{)}^3}$

The total number of books is $a+2b+3c+d$. The total number of ways in
which these books can be arranged in a shelf (in same row) is
$=\frac{(a+2b+3+d)!}{a!(b!{)}^2(c!{)}^3}$